获取HttpRequest.responsexml null

Question

我正在尝试获取rest API的响应，但它返回null。 这是我的JS代码：

function submitApproval()
{
    //alert('Approval Start');
    var invocation = new XMLHttpRequest();
    //alert('Get Url');
    var url ='http://localhost:port/rest/bpm/wle/v1/task/303?parts=all';
    send_with_ajax(url);
}
function send_with_ajax(url){
    //  alert('Send Start'+url);
    var httpRequest = new XMLHttpRequest();
    //alert("Start");
    httpRequest.open("GET", url, true);
    httpRequest.setRequestHeader("Content-type", "application/xml");
    httpRequest.setRequestHeader("X-Requested-With", "XMLHttpRequest");
    httpRequest.setRequestHeader("X-Alt-Referer", "http://www.google.com");
    alert("Send");
    httpRequest.send();
    alert('Send Request');
    var _tempRecommendations = httpRequest.responseXML;
    // window.alert(httpRequest.response)
    //window.alert(httpRequest.responseText)
    window.alert(_tempRecommendations);
    /*    
    httpRequest.onreadystatechange = function() { 
        alert('onLoad call');
        if (httpRequest.readyState == 4)
        {
            var _tempRecommendations = httpRequest.responseXML;
            window.alert(httpRequest.response)
            window.alert(httpRequest.responseText)
            window.alert(_tempRecommendations)
        }
    };*/  
}

我该怎么做？我尝试过onload和onreadystatechange但结果相同。

Answer 1

您应该在设置onreadystate处理程序之后发送请求，以确保在收到响应之前设置处理程序。

// set this first
httpRequest.onreadystatechange = function() { 
    window.alert(httpRequest.responseText);
};

// and then
httpRequest.send();

Answer 2

Here is working Demo。检查readystate事件

中的onreadstatechange

 function test(){
    var xhr = new XMLHttpRequest();
    xhr.onreadystatechange = function () {
        if (xhr.readyState == 4) {
            alert(xhr.responseText);
        }
    }
    xhr.open('GET', 'http://example.com', true);
    xhr.send(null);
}

test();